a
Reaction suggests that $1$ mole of aniline give $1$ mole of orange dye.
$\text { so }(\mathrm{mol})_{\text {aniline }}=(\mathrm{mole})_{\text {orange dye }}$
$\frac{9.3 \mathrm{~g}}{93 \mathrm{~g} \mathrm{~mol}^{-1}}=\frac{\text { mass of orange dye }}{199 \mathrm{~g} \mathrm{~mol}^{-1}}$
$\text { mass of orange dye }=19.9 \mathrm{~g} \simeq 20 \mathrm{~g}$
